# What is the arclength of f(t) = (-2(t+3)^3,(t-2)^2) on t in [1,2]?

##### 1 Answer
Jun 3, 2018

$\approx 122.006$

#### Explanation:

We have
$x \left(t\right) = - 2 {\left(t + 3\right)}^{3}$
so
x'(t)=-6)t+3)^2
$y \left(t\right) = {\left(t - 2\right)}^{2}$
so

$y ' \left(t\right) = 2 \left(t - 2\right)$

so our integral is given by

${\int}_{1}^{2} \sqrt{{\left(- 6 {\left(t + 3\right)}^{2}\right)}^{2} + {\left(2 \left(t - 2\right)\right)}^{2}} \mathrm{dt}$
A numerical method gives us $\approx 122.006$